A car covers AB distance with first one – third at velocity $V_{1}m{s}^{-1}$ second one-third at $V_{2}m{s}^{-1}$ and last one-third at $V_{3}m{s}^{-1}.$ If $V_3=3V_1,V_2=2V_1$ and $V_1=11m{s}^{-1}$ then the average velocity of the car is __ $m{s}^{-1}$

 

Given situation is now, Let total displacement = x 
Then, time to cover first $\frac{x}{3}$

$\begin{array}{l}displacement,t_1=\frac{displacement}{velocity}\Rightarrow t_1=\frac{x}{3X11}\\ Similarly,t_2=\frac{x}{3X22}and{t}_3=\frac{x}{3X33}\\ Now,V_{average}=\frac{Totaldisplacement}{Totaltime}=\frac{x}{\frac{x}{33}+\frac{x}{66}+\frac{x}{99}}\\ 1=\frac{1}{\frac{1}{33}(1+\frac{1}{2}+\frac{1}{3})}=\frac{33}{\left(\frac{6+3+2}{6}\right)}=\frac{33X6}{11}=18m/s\end{array}$